10.3 - The nth Term Test for Divergence

Introduction

The nth term test for divergence (also called the divergence test) is a quick check that can sometimes prove a series diverges.

The basic idea: if the individual terms of a series don’t shrink toward zero, there’s no way the sum can settle on a finite value.

Statement of the Test

For

\sum_{n=1}^{\infty} a_n

If the series converges, then

\lim_{n \to \infty} a_n = 0

The contrapositive is the useful version:

\text{If } \lim_{n \to \infty} a_n \neq 0 \text{ (or doesn't exist), then } \sum_{n=1}^{\infty} a_n \text{ diverges}

Critical Warning: This Test Only Goes One Way

What it CAN tell you: if

\lim_{n \to \infty} a_n \neq 0

, the series diverges.

What it CANNOT tell you: if

\lim_{n \to \infty} a_n = 0

, you get no conclusion. The series might converge, or it might diverge.

The Trap

\lim_{n \to \infty} a_n = 0 \quad \not\Rightarrow \quad \text{the series converges}

That’s bad reasoning. When the limit is zero, you need a different test.

Using the Test

Identify $a_n$ , the general term

Calculate $\lim_{n \to \infty} a_n$

Decide:

Limit $\neq 0$ means the series diverges (say “by the nth term test”)

Limit $= 0$ means no conclusion, so reach for another test

Example 1: Clear Divergence

Does

\sum_{n=1}^{\infty} \frac{4n^2 + 3n + 1}{7n^2 - 2}

converge or diverge?

For a rational function, divide numerator and denominator by the highest power of

n

\lim_{n \to \infty} \frac{4n^2 + 3n + 1}{7n^2 - 2} = \lim_{n \to \infty} \frac{4 + \frac{3}{n} + \frac{1}{n^2}}{7 - \frac{2}{n^2}} = \frac{4}{7}

Since

\frac{4}{7} \neq 0

, the series diverges by the nth term test.

Example 2: When the Test Gives Nothing

Does

\sum_{n=0}^{\infty} 5\left(\frac{1}{3}\right)^n

converge or diverge?

\lim_{n \to \infty} 5\left(\frac{1}{3}\right)^n = 5 \cdot 0 = 0

The nth term test gives no conclusion.

But we can recognize this as a geometric series with

|r| = \frac{1}{3} < 1

, so it converges, and

\sum_{n=0}^{\infty} 5\left(\frac{1}{3}\right)^n = \frac{5}{1 - \frac{1}{3}} = \frac{5}{\frac{2}{3}} = \frac{15}{2}

Key point: the nth term test was useless here, but geometric series knowledge wasn’t.

Example 3: Terms Go to Zero, Still No Conclusion

Does

\sum_{n=1}^{\infty} \frac{1}{n^2 + 5}

converge or diverge?

\lim_{n \to \infty} \frac{1}{n^2 + 5} = 0

The nth term test gives no conclusion. (This series does actually converge, but you’d need a different test to prove it.)

Example 4: Simplify Before Taking the Limit

Does

\sum_{n=1}^{\infty} \frac{3^{n+1}}{3^n \cdot 5 + 1}

converge or diverge?

Step 1: Simplify. Since

3^{n+1} = 3 \cdot 3^n

a_n = \frac{3 \cdot 3^n}{5 \cdot 3^n + 1}

Step 2: Take the limit. For large

n

, the

+1

is negligible next to

5 \cdot 3^n

\lim_{n \to \infty} \frac{3 \cdot 3^n}{5 \cdot 3^n + 1} = \frac{3}{5}

Since

\frac{3}{5} \neq 0

, the series diverges by the nth term test.

Example 5: Exponential Growth

Does

\sum_{n=1}^{\infty} \frac{e^{5n}}{4n}

converge or diverge?

This is

\frac{\infty}{\infty}

, so use L’Hôpital’s Rule:

\lim_{n \to \infty} \frac{e^{5n}}{4n} = \lim_{n \to \infty} \frac{5e^{5n}}{4} = \infty

Since the limit isn’t zero, the series diverges by the nth term test.

Exponentials in the numerator crush polynomials in the denominator.

Example 6: Oscillating Terms

Does

\sum_{n=1}^{\infty} (-1)^n

converge or diverge?

\lim_{n \to \infty} (-1)^n \text{ does not exist}

The terms bounce between

-1

(odd

n

) and

1

(even

n

) forever.

Since the limit doesn’t exist (and so certainly isn’t zero), the series diverges by the nth term test.

Common Limit Techniques

Rational functions. For

\lim_{n \to \infty} \frac{P(n)}{Q(n)}

with polynomials

P

and

Q

: divide both by the highest power of

n

, then compare leading coefficients.

Exponential vs. polynomial. Exponential growth wins:

\lim_{n \to \infty} \frac{e^{an}}{n^k} = \infty \text{ for any } a > 0 \text{ and any } k

L’Hôpital’s Rule. When you hit

\frac{\infty}{\infty}

\frac{0}{0}

\lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{f'(n)}{g'(n)}

You may need to apply it more than once.

Why the Test Fails When the Limit Is Zero

Intuition: just because the individual terms get small doesn’t mean you aren’t adding infinitely many of them.

Classic counterexample: the harmonic series.

\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

Here

\lim_{n \to \infty} \frac{1}{n} = 0

, but the series diverges. The terms approach zero, just not fast enough to keep the total finite.

Summary Table

Condition	Conclusion	Why
$\lim_{n \to \infty} a_n \neq 0$	Diverges	Terms aren’t approaching zero
$\lim_{n \to \infty} a_n$ doesn’t exist	Diverges	Terms aren’t approaching zero
$\lim_{n \to \infty} a_n = 0$	Inconclusive	Need another test

When to Use It

Always try the nth term test first, because it’s fast, and if it shows divergence you’re done. If it comes back inconclusive, at least you know you need something else.

Best for:

Rational functions where numerator and denominator have the same degree

Series with exponential terms

Obviously divergent series

Useless for:

Series where the terms clearly go to zero

Key Takeaways

nth term test: if $\lim_{n \to \infty} a_n \neq 0$ , then $\sum a_n$ diverges

One-way test: it can prove divergence but never convergence

Limit equals zero means no conclusion

Always try it first, it’s the quickest route to a divergence answer

Common error: assuming $\lim a_n = 0$ implies convergence. It doesn’t.

Geometric series can still be settled with $|r| < 1$ even when this test is inconclusive

State the conclusion as “the series diverges by the nth term test”

L’Hôpital’s Rule is often handy for the limit

If the limit doesn’t exist (like an oscillation), the series diverges

When $\lim a_n = 0$ , you need the tests from the next lessons

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